# Cannot determine the P-value [closed]

I have a chi squared of 0.1185 with 1 degree of freedom. How can i determine the P-value?

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Because $\chi^2(1)$ is the distribution of the square of a standard normal variate, your question is tantamount to asking for the probability that a standard normal value exceeds $\sqrt{0.1185}\approx0.34$ in size. That's an elementary calculation (and I'm sure you have software or accurate tables to carry it out). – whuber Mar 5 '12 at 23:32

## closed as not a real question by whuber♦Aug 14 '12 at 12:38

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, see the FAQ.

By p-value I presume you mean "the probability of seeing a value of my test statistic as large as this, if it really comes from a $\chi^2$ distribution with one degree of freedom". So if you get a low p-value it means it is an improbable value to have seen.

In R:

1-pchisq(0.1185, 1)


or

pchisq(0.1185, 1, lower=FALSE)


In Excel

=1-CHISQ.DIST(0.1185,1, TRUE)


or

=CHISQ.DIST.RT(A2,1)


The p-value is very large because your value of the test statistic is actually considerably less than the average value of such a $\chi^2$ random variable. If anything, it is unusually small. So basically, you do not have a large value and hence there is not evidence to dismiss the hypothesis that it genuinely comes from a $\chi^2$ distribution.

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so, what is the p-value? – user176105 Mar 5 '12 at 22:54
I would recommend you always use FALSE instead of F, so it should be pchisq(0.1185, 1, lower = FALSE). – Henrik Mar 5 '12 at 23:08
@user176105 I'll leave it up to you to calculate... you must have access to either R or Excel, surely. – Peter Ellis Mar 5 '12 at 23:10
@PeterEllis so my teacher just gave me a table like this: sociology.ohio-state.edu/people/ptv/publications/p%20values/… so the only thing i can do is narrow my p-value to a range? – user176105 Mar 5 '12 at 23:14
Linear interpolation of the 2-tailed p-value for $z=\sqrt{0.1185}$ gives $0.73$ which is fully accurate to two d.p. (In R this can be done as (sqrt(0.1185)-.39)/(.25-.39) * (.80-.70) + 0.70; the numbers 0.39, 0.25, 0.80, and 0.70 come from your teacher's table.) Alternatively, applying Simpson's Rule (x<-0.1185;1 - sqrt(x)*(2*exp(-x/2) + 4)/(3*sqrt(2*pi))) gives 0.7306, which is within 0.0001 of the truth: no table is needed at all. – whuber Mar 5 '12 at 23:43
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The upper tail of the $\chi^2_1$ distribution, beyond 0.1185 is

pchisq(0.1185, df=1, lower=F)
[1] 0.7306671


@PeterFlom Because only TRUE and FALSE are reserved keywords (see ?reserved). F and T are just variables assigned to your convenience. So you can easily use the following construction to produce horrible things: T <- FALSE; F <- TRUE. – Henrik Mar 6 '12 at 7:41